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seen Mar 10 '13 at 15:31

My name is Paul Le Meur. I live in Bordeaux, France.


Sep
24
awarded  Autobiographer
Nov
11
awarded  Yearling
Dec
16
asked Hilbert transform and Hilbert matrix
Dec
5
revised Conditional convergence of Riemann's $\zeta$'s series
added 383 characters in body
Dec
5
revised Conditional convergence of Riemann's $\zeta$'s series
added 1047 characters in body
Dec
5
comment Conditional convergence of Riemann's $\zeta$'s series
Well it does not explain things very clearly but I could find that $\zeta$ does not converge conditionally at the left of the abscissa of absolute convergence. It does not address points $1+it$, $t\ne 0$.
Dec
5
comment Conditional convergence of Riemann's $\zeta$'s series
The first result in that link, on uniform convergence, is really tight, it implies that there is no pole on the real half-line at the right of the abscissa of conditional convergence, for any Dirichlet series. So $\zeta$ does not converge conditionally for $\sigma<1$. That does not settle $\sigma=1$. (I am going to read other answers.)
Dec
4
comment Conditional convergence of Riemann's $\zeta$'s series
Great paper (by You Tong Siu I remark), I do not have time to read it much now, but it does seem that $\zeta$ converges on and at the left of its "abscissa of convergence".
Dec
4
comment Conditional convergence of Riemann's $\zeta$'s series
I do not think that the abscissa of convergence refers to issues of conditional convergence neither that it implies much (straightforwardly) about my question. As mentioned above it seems that uniform convergence of the partial sums may occur at the left of the abscissa of (absolute) convergence. Thanks for the attempt though.
Dec
4
revised Conditional convergence of Riemann's $\zeta$'s series
added 472 characters in body
Dec
4
comment Conditional convergence of Riemann's $\zeta$'s series
The page you point to does not address my question. Your answer only treats absolute convergenge, which I understand. The wiki page says: "absolute convergence and uniform convergence may occur in distinct half-planes" which I guess implies that Dirichlet series may converge conditionally on the left of their "abscissa of convergence" (which really refers to absolute convergence). My question remains: do the partial sums converge conditionally for some $\sigma=1$?
Dec
4
comment Conditional convergence of Riemann's $\zeta$'s series
Your implications (assuming you meant $|n^s|$ where there is a typo) only show the sum does not converge absolutely, not that it does not converge conditionally.
Dec
4
awarded  Student
Dec
4
asked Conditional convergence of Riemann's $\zeta$'s series
Nov
11
awarded  Yearling
May
30
answered Function bounded below, with negative 2nd derivative. Show function is constant.
May
30
comment Function bounded below, with negative 2nd derivative. Show function is constant.
The intuitive picture: Assume $f(x)>l$, $f'$ is nonincreasing so if it is not $0$ at some point $x_0$, take the tangent to the graph of $f$ there. Since this line has nonzero slope if will cut the lower bound line $y=l$ either on the left or on the right of $x_0$, and the graph of $f$ lies below that tangent, so it will pass below $y=l$ too, therefore $f'$ vanishes on $\mathbb R$.
May
26
comment Functional Analysis- Convergence
I mean, we assume the matrix is self-adjoint, hermitian. These correspond to hermitian inner products for positive matrices, or "semi-inner product" for nonnegative ones.
May
26
comment Functional Analysis- Convergence
Nonnegative is that $\langle Tx,x\rangle\ge 0$ for all $x$. Over $\mathbb C$, in finite dimension, you'll have diagonalizable matrices with diagonal forms with positive or zero entries. Positive is then when none in the diagonal is zero.
May
24
comment Explaining why sin and cos are *not* at right angles
I think the answer is yes: being a vector on the real axis and being a scalar is the same, this is because $\mathbb C$ is not just a vector space over $\mathbb R$ but a field extension. So that $1$ is a very special vector in $\mathbb C$ (as a real plane), it is its identity as a field. The subspace spanned by $1$ over $\mathbb R$ is then distiguished: it is both the "real axis" and the "scalars" -if you just think of $\mathbb C$ as a real plane.